Description
We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change
it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions.
1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2).
2. Q x y (1 <= x, y <= n) querys A[x, y].
Input
The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2
y2", which has been described above.
Output
There is a blank line between every two continuous test cases.
Sample Input
1 2 10 C 2 1 2 2 Q 2 2 C 2 1 2 1 Q 1 1 C 1 1 2 1 C 1 2 1 2 C 1 1 2 2 Q 1 1 C 1 1 2 1 Q 2 1
Sample Output
1 0 0 1
Source
很好的题啊,一开始想错更新方向了,怎么都想不通.....
求某个点的值就是求这个点的更新次数,因为这个矩阵是01矩阵
而求这个点的更新次数又可以转化为求树状数组子矩阵 (1,1)---(x,y)的和,可以这么想,所有会让点(x,y)修改的点都在他左上方的矩阵上,那么那些点每更新一次,点(x,y)也要更新一次
#include<stack> #include<queue> #include<vector> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; const int N=1010; int tree[N][N]; int lowbit(int x) { return x&(-x); } void add(int x,int y) { for (int i=x;i<=N;i+=lowbit(i)) for (int j=y;j<=N;j+=lowbit(j)) tree[i][j]++; } int sum(int x,int y) { int ans=0; for (int i=x;i;i-=lowbit(i)) for (int j=y;j;j-=lowbit(j)) ans+=tree[i][j]; return ans; } int main() { int x,n,T,x1,y1,x2,y2; char op[3]; scanf("%d",&x); while (x--) { scanf("%d%d",&n,&T); memset(tree,0,sizeof(tree)); while (T--) { scanf("%s",op); if (op[0]=='C') { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); add(x1,y1); add(x1,y2+1); add(x2+1,y1); add(x2+1,y2+1); } else { scanf("%d%d",&x1,&y1); printf("%d\n",sum(x1,y1)%2); } } printf("\n"); } return 0; }